Wiring RESISTORS for Firebox LED (12V DC Analogue)

[70021 Morning Star]

.

THIS QUESTION IS ABOUT RESISTORS

 

Surely I can't be alone in wanting to add a flickering LED firebox light and a smoke-generator to an old style 12V (analogue) locomotive?

 

I DO NOT want to add a DCC Chip, as it's a waste of money when I only need the correct resistor costing a few pence; but, I would also like to fit a suitable resistor to proof against the higher (15V) power on DCC layouts, if I should mistakenly put them onto a DCC layout..

 

I've already sourced the appropriate 3mm flickering LEDs suitable for the glow of the firebox. They're 3V LEDs. But there I'm running into problems.

 

I've got a simple power take off from the DC Hornby locomotive. 12V DC. But, I'm totally confused about what resistor I need to use to protect the 3V LED. ...according to what I've read, the calculation is Available Power (12V) minus the LED Power Rating (3V), which gives us 9V. This is divided by the 0.015, representing the diode 'forward current'. This says I need to use  680Kohms resistor.

 

Problem is that I'm reading people are using 1K (1,000Kohms) resistors. While my calculation suggests that a 3V LED on a 12V circuit needs a 680K (680,000ohms) resistor, which doesn't sound right.

 

And, if I want to protect my 3V LED in case I put the locomotive on a 15V DCC circuit, the numbers get even worse...

 

Surely this is something that people did many times before the introduction of DCC ? It must be a pretty standard answer for which resistor to use on a 12V analogue circuit with a 3V LED, but I can't find the answer anywhere. 

 

Similarly, wiring a smoke-generator will probably require a resistor.

 

HELP PLEASE

 

Thanks,

Rick

[tractor_37260]

.

THIS QUESTION IS ABOUT RESISTORS

 

Surely I can't be alone in wanting to add a flickering LED firebox light and a smoke-generator to an old style 12V (analogue) locomotive?

 

I DO NOT want to add a DCC Chip, as it's a waste of money when I only need the correct resistor costing a few pence; but, I would also like to fit a suitable resistor to proof against the higher (15V) power on DCC layouts, if I should mistakenly put them onto a DCC layout..

 

I've already sourced the appropriate 3mm flickering LEDs suitable for the glow of the firebox. They're 3V LEDs. But there I'm running into problems.

 

I've got a simple power take off from the DC Hornby locomotive. 12V DC. But, I'm totally confused about what resistor I need to use to protect the 3V LED. ...according to what I've read, the calculation is Available Power (12V) minus the LED Power Rating (3V), which gives us 9V. This is divided by the 0.015, representing the diode 'forward current'. This says I need to use  680Kohms resistor.

 

Problem is that I'm reading people are using 1K (1,000Kohms) resistors. While my calculation suggests that a 3V LED on a 12V circuit needs a 680K (680,000ohms) resistor, which doesn't sound right.

 

And, if I want to protect my 3V LED in case I put the locomotive on a 15V DCC circuit, the numbers get even worse...

 

Surely this is something that people did many times before the introduction of DCC ? It must be a pretty standard answer for which resistor to use on a 12V analogue circuit with a 3V LED, but I can't find the answer anywhere. 

 

Similarly, wiring a smoke-generator will probably require a resistor.

 

HELP PLEASE

 

Thanks,

Rick

 

Rick

      Either 680ohm (NOT 680K) or a 1K resistor could be used. If possible use on a DCC layout is planned use the 1K option. Track voltages on DCC are not always 15V , they can be higher depending on what system is used.  

 

Trial using both resistor options on DC, the 680ohm will probably be slightly brighter relative to the speed control setting than the 1K

 

When using DC power - remember the brightness of the LED will also vary as you turn the DC controller up/down and it will also go off altogether when the loco stops. 

 

One of the many positives of DCC operation - lights can "stay" on when locos are stopped.

 

Perhaps someone else will come along and clue you up on using smoke generators with DC ?

 

HTH

Ken

[70021 Morning Star]

Thanks Ken,

The plan is for my local club to upgrade to DCC at "some time", but I think it'll be for a year or two before things are up and running, otherwise life would be so simple.

 

Meanwhile, thanks to your assistance, I've now ordered:

Carbon Film Resistors 680R ohm and 1K ohm in a quantity that'll last me a lifetime!

 

Only issue I can foresee is that the LED will only work with the locomotive under power in the forwards direction, as the +ve/-ve power pickup won't feed the correct polarity to the LED when travelling in reverse. I'm guessing the only solution would be to use two LEDs, one wired +ve and one wired -ve?

 

Cheers,

Rick

[smokebox]

You could fit a bridge rectifier to allow the firebox flicker to work in both directions.

[cobach47]

you will need two l.e.d's as they are polarity sensitive  give me your address i'll put some resistors in the post . the resistor is soldered on the longest lead then you connect

 

the two together but with reversed connections then connect to power supply it will work in both directions  -   cheers Ian

[70021 Morning Star]

Smokebox,

A 'bridge rectifier'. Excellent, but I've no idea what that is, although I've heard the name somewhere.

 

Can you please explain, in simple terms, what a 'bridge rectifier' is and how I would go about obtaining and fitting one?

 

Meanwhile,

The LED supplier has furnished the technical specifications (from the manufacturer), so I've been able to do the correct calculations for determining what resistor is needed. I'll come back to this in a separate posting.  

 

Cobach47,

That's very generous of you. Thank you. I'll post the corrected calculations - based on the manufacturer's data - then get back to you.

 

Smoke-generator

This appears relatively straightforward. Purchase a suitable 12V or 12-16V generator, and wire directly to the power pickup. We're talking about DC analogue - not DCC digital - of course. Only requirement is to control the heat - I'll probably fit a 'heat sink', as this is a tender-drive, and there's plenty of room inside the locomotive - as well as a resistor, to ensure the thing isn't going full blast all of the time. I may also fit a micro switch, so I can turn the smoke-generator off when I don't want to use it. ...I'm still looking for a suitable way of powering the smoke-generator off. Any suggestions?

 

I've got a very old 1970s? locomotive that uses an interesting system of smoke generation without melting the plastic. It has a huge heat-sink, and I'm thinking this principle has possibilities. Anyone have any thoughts here? I'd also like to leak smoke around the pistons and safety-valves. Anyone ever tried this?

 

Anyway,

My hope is that we can post a definitive guide for others who hope to fit a firebox LED and/or a smoke-generator to DC analogue locomotives, without the pain I'm going through. ...thankfully, there's guys like you, on RMWEB, to help us out!

 

Thanks, everyone,

Rick

[RJS1977]

A bridge rectifier is an arrangement of four diodes that can be connected across an AC or reversible DC supply to give a constant polarity DC. They're also used in DC controllers (indeed any mains-powered gadget requiring a DC supply) to convert the low-voltage AC supply from the transformer to DC which may be where you've heard the term before.

 

Maplin's sell them (though past experience with Maplin's is that they always have one fewer of any component than you need...).

 

Online stores like Cricklewood Electronics sell them too but they're so cheap most of what you'd pay would be postage!

 

http://www.cricklewoodelectronics.com/product.php?productid=8963&page=1

[70021 Morning Star]

Hi RJS1977,

25p sounds good. I'll be needing two, but even 50p is within budget :-)

 

Only question is what do I do with them?

 

My current wiring diagram couldn't be simpler. +ve and -ve take offs from the polarized (left/right or forward/reverse) track pickupsthat normally feed the DC electric motor. I tap into these, to provide wires to the LED, with a resistor on the +ve feed. But, how is the bridge rectifier wired into this?

 

Also, would you have a Maplin part number, please?

 

Thanks,

Rick  

[smokebox]

 

My current wiring diagram couldn't be simpler. +ve and -ve take offs from the polarized (left/right or forward/reverse) track pickupsthat normally feed the DC electric motor. I tap into these, to provide wires to the LED, with a resistor on the +ve feed. But, how is the bridge rectifier wired into this?

 

Also, would you have a Maplin part number, please?

 

Thanks,

Rick

 

Take the +ve and -ve feeds to the inputs of the bridge rectifier, often marked ~ or ac. You then connect the LED and series resistor to the +ve and -ve outputs of the bridge rectifier.

 

Maplin part number AQ98G would be OK if they have any in stock. This has a 2 amp capacity which is much higher than necessary but appears to be the smallest that Maplin have listed at the moment but they seem to be removing it from their list.

[RJS1977]

70021

 

Maplin part no is W005:

 

http://www.maplin.co.uk/p/w005-15a-silicon-bridge-rectifier-aq94c

 

On the casing, two leads will be marked with a ~ - these go to the AC supply, or in this case, the track pickups. The other two are marked + and - and go to the resistor/LED.

 

Richard

[Il Grifone]

871

This is the sort of thing required, (unless you are in a hurry - supplies from Hong Kong can take a while).

 

http://www.ebay.co.uk/itm/10PCS-DB107-DIP-4-DB-107-1000V-1A-BRIDGE-RECTIFIER-/221516788461?pt=LH_DefaultDomain_3&hash=item33936df2ed

 

The bridge can also be made up from 4 single diodes.

 

That it goes out sometimes is not really a problem. The firebox door(s) would only be open while the locomotive was being fired.

912

915

[70021 Morning Star]

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CALCULATIONS

 

Thanks Guys,

Now let's begin by checking I've done the right calculations.

 

To obtain the required resistance, it's Supply Voltage minus LED Voltage Consumption, and divide the result by the LED's (Continuous) Forward Current.

In my case the numbers I want to use are 12V and 16V for DC and DCC

minus 3.3V consumed by the LED

divided by the LED's 35mA forward current.

12V - 3.3V = 8.7V

8.7V / 0.035A = 248.571 ohms

or

16V - 3.3V = 12.7V

12.7V / 0.035A = 362.857 ohms

Which, rounding up, says I need a 249 ohms resistor for 12V DC or a 363 ohms resistor for 16V DCC.

Hopefully, the flickering LED will be sufficiently bright, and its flicker circuitry will work with the DCC protecting 363 ohms, but I'd best try both.

 

As for 'worst case' Heat Dissipation (expresses in Watts). I understand this is Supply Voltage multiplied by Supply Voltage, divided by the Power (in Amps).

So, for my 16V DCC Protection, that's 16V multiplied by 16V, divided by the 363 ohms resistor calculated above.

16V x 16V = 256V

256V / 363A =  0.705W

or, for DC only

12V x 12V = 144V

144V / 249A = 0.578W

 

So I'm thinking I need some 1W rated 390R 363 ohms (and, probably, 270R 249 ohms) resistors. Is this correct?

 

Thanks,

Rick

[kevinlms]

.

CALCULATIONS

 

Thanks Guys,

Now let's begin by checking I've done the right calculations.

 

To obtain the required resistance, it's Supply Voltage minus LED Voltage Consumption, and divide the result by the LED's (Continuous) Forward Current.

In my case the numbers I want to use are 12V and 16V for DC and DCC

minus 3.3V consumed by the LED

divided by the LED's 35mA forward current.

12V - 3.3fV = 8.7V

8.7V / 0.035A = 248.571 ohms

or

16V - 3.3fV = 12.7V

12.7V / 0.035A = 362.857 ohms

Which, rounding up, says I need a 249 ohms resistor for 12V DC or a 363 ohms resistor for 16V DCC.

Hopefully, the flickering LED will be sufficiently bright, and its flicker circuitry will work with the DCC protecting 363 ohms, but I'd best try both.

 

As for 'worst case' Heat Dissipation (expresses in Watts). I understand this is Supply Voltage multiplied by Supply Voltage, divided by the Power (in Amps).

So, for my 16V DCC Protection, that's 16V multiplied by 16V, divided by the 363 ohms resistor calculated above.

16V x 16V = 256V

256V / 363A =  0.705W

or, for DC only

12V x 12V = 144V

144V / 249A = 0.578W

 

So I'm thinking I need some 1W rated 363 ohms (and, probably, 249 ohms) resistors. Is this correct?

 

Thanks,

Rick

No, not exactly. Resistors come in what's termed the E12 or E24 range, which means there are some common values in a Decade (either 12 or 24).

 

So for E12, the values between 1 & 10 are 1.0, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8 & 8.2.

 

Between 10 & 100, the values are multiplied by 10 and between 100 & 1000, times 100, etc.

 

 

The closest resistors you'll get to your calculated ones EASILY & CHEAPLY are 390 ohms in the E24 series and 270 Ohms, also in the E24 series. (ALWAYS use a higher, rather than lower value).

 

I haven't checked your calculations.

 

Edited to delete link, which isn't happy.

[70021 Morning Star]

Thanks Everyone,

I'm starting to understand this now.

 

Yes, Kevin, I'd just put the raw results of my calculation, but had updated my post with 390R and 270R codes while you were writing your reply.

 

I can only get these in 2W form, but take it that just means they're bigger than the 1W version; as well as giving more heat protection?

 

Just about to drive over to Maplin to collect the resistors and bridge rectifiers. Then I'll need to study Smokebox's instructions...

 

Cheers,

Rick

[smokebox]

I wouldn't use the quoted forward current, that's the maximum that the LED can withstand. It is far better to use a lower value, maybe about two thirds of the maximum quoted, so 20mA would be my preference. This is normally bright enough and often more than bright enough with most LEDs and helps to extend it's life.

[70021 Morning Star]

Good thinking, Smokebox,

I'm thinking to use the 390R on the 12V DC, which is overrated to protect in case I forgetfully put her on a 16V DCC circuit, so should be ok. I'd actually taken the manufacturer's lower rating and already dropped it a smidgen, it's time to blow a few circuits.  I've got three spare LEDs, so can afford a couple of mishaps.

 

If I use your suggested 20mA, that gives 635 ohms, which I've already got. So will take your advice and work backwards from higher to lower.

  

Cheers,

Rick

[Il Grifone]

1081

LEDs are fairly constant brightness devices independent of current. I would set the current at 10mA, which requires a 1 kilohm resistor (there's no need to be precise). If the brightness is adequate even less would be preferable - try  a couple of resistors in series to make a higher resistance.

 

It's rating is given by the square of the voltage divided by the resistance*. Taking the full supply voltage  (to err on the safe side) this is 15 X 15 / 1000 or 225 milliwatt. A 1/4 watt resistor is thus adequate. A higher power rating is perfectly OK, but will take up more space.

 

* P = V x I  and V = R x I

[70021 Morning Star]

Hello David,
What you say about LEDs being, essentially, a binary on/off device, as versus the analogue brightness increase of incandescent bulbs. So trying  1K resistor sounds good.
 
Only problem is that I haven't got any 1K resistors. I could link a couple in series, but they're not rated to 1 Watt. So I have to ask, do two 1/2K resistors linked in series reduce the Wattage power,  or will one just burn out?
 
Also, as these are "flickering" LEDs, they have some circuitry inside the bulb. I don't know if that affects anything, but here's the manufacturer's full specification (just received from the seller):

3.2-3.4V (CE&Rosh)
Material: AlGaInP
Absolute Maximum Rating at Ta = 25°C
Power Dissipation: 100 mW
Peak Forward Current: 100mA
(1/10 Duty.Cycle. 0.1ms Pulse width)
Continuous Forward Current: 35mA
Derating Linear From 50°C : 0.4mA/°C
Reverse Voltage: 5V
Operating Temperature Range: -40°C to +80°C
Storage Temperature Range: -40°C to +80°C
Lead Soldering Temperature: 260°C for 5 Seconds

 
Thanks again,
Rick

[Il Grifone]

1288

Hello Rick,

 

Increasing the number of resistors will increase the power handling capacity. For example, two 1/2W resistors of (say) 1 kilohm can be connected in series to give 2 kilohm or in parallel to give 500 ohms. In both cases the the power handling increases to 1W*. (If the resistors are of different values, the maths becomes more complicated.) There is a tolerance on the resistor value of course, usually 10% (silver band) or 5% (gold band).

 

* The rating assumes the component is in free air. Mounting them close together will reduce the theoretical rating.

 

LEDs require a reverse biased diode in parallel for protection if fed from AC (it can be another LED or the bridge rectifier feed will do instead), but saying that I've never blown one up by leaving it out. The flickering device may be more sensitive, however, so I wouldn't chance it.

 

The flickering effect can be produced by a music chip (similar to the ones in greeting cards etc.).

 

Regards

1298

[70021 Morning Star]

Hi 1288,

Sorry, you misunderstand my question. I'm asking about Watts, not Ohms.

 

I've got a 0.7W power supply. That's too much for a 250R ohm 1/8W 0.125W carbon film resistor to handle. I'd need to raise the 0.125W to 0.7W minimum, yes?

 

So, if I linked 6 of the 0.125W resistors in series (totalling 0.75W), would that make them share the Wattage, or simply burn out the first resistor? (We're ignoring the ohms for a moment.) 

 

Thanks,

Rick

[smokebox]

The power rating of the supply, 0.7 Watts, is the maximum that it could supply. It will be happy supplying any amount up to that figure.

[Il Grifone]

914

Hi,

 

Basically the resistance sets the current and hence the power consumed for a given voltage. For a 12V supply, the current is limited to 48mA through a 250 ohm resistor, which is 0.6W. Doubling the resistance will reduce the current drawn by half and the power consumed to a quarter.

 

Assuming all the resistors are the same value, they will indeed share the load equally. In the event of an overload, they will all heat up and eventually one will fail, though all would be damaged in this circumstance.

 

It is good practice to allow a worst case margin on any calculations. There is a tolerance in the resistor value and cheap power supplies are notorious for poor regulation. The voltage quoted is at rated power (or should be), but can rise considerably off load - possibly 50% (or more). A cheap test meter is useful for experimentation and can help with tracking down faults on the layout.

 

Smoke generators are designed for 12V so shouldn't need a resistor (over production of smoke was not a problem with the few I've had!).

 

Regards,

924

[70021 Morning Star]

Ok, I'm clear on Resistance (ohms) calculations for powering LEDs. I also understand how to calculate the Wattage (W)of the required resistors. So we're good so far.

 

But I've still not got an answer on using multiple resistors. I've read the math on the Internet, but want to verify I've understood correctly. So, let me run it by you...

 

If I've calculated I need a 363 ohms resistor on a 0.7 Watt circuit. I find the next highest resistor that's available (a 390R), with a slightly higher Wattage. So, a 390R rated at 1W appears appropriate.

 

Problem is sourcing the 390R at 1W. Maplins only sell 2W and higher. While ebay sellers are mostly selling lower Wattages, 1/4W (0.25W) or 1/8W (0.125W). My question is, can create a 390 ohm resistance - and get the correct Wattage -  using smaller Wattage resistors linked in series?

 

            for example: 3x 1/4W resistors = 0.75W (which is higher than my 0.7W power supply).

 

Of course, these resistors would need to be 1/3rd of the 390 ohms, as the resistance is cumulative in a serial circuit. So 390 / 3 = 130.

 

If I can't obtain a 1W 390ohms resistor, I could substitute with 3x 1/4W 130ohm resistors (YES/NO)?

 

Thanks,

Rick

[70021 Morning Star]

.

FIBRE OPTIC LAMPS

 

Slightly 'off topic', but I've been looking at using the Firebox flickering-LED to illuminate the locomotive's headlamps and the tender tail lamp. As oil lamps actually burn yellow, and flicker, I think this will look more realistic than those very expensive miniature LED locomotive oil lamps, and at a fraction of the cost. As for the tender's red tail light, that's easily coloured Red with one of Staedtler's  permanent OHP pens.

 

Anyway, there's very good instructions on fitting fibre-optics elsewhere on RMWEB. However, on thing that's not covered is the shape of the fibre ends. Reading up on fibre-optic 'splicing', it seems that rounding the ends of the fibre - even in butt-joints - will give the best light transference. The fibre-optic experts recommend an 8 degree angle. ...just thought I'd mention it, as it requires planning into the Firebox LED installation.

 

I'll post some results later (probably much later ;-).

 

Rick

[RJS1977]

 

Anyway, there's very good instructions on fitting fibre-optics elsewhere on RMWEB. However, on thing that's not covered is the shape of the fibre ends. Reading up on fibre-optic 'splicing', it seems that rounding the ends of the fibre - even in butt-joints - will give the best light transference. The fibre-optic experts recommend an 8 degree angle. ...just thought I'd mention it, as it requires planning into the Firebox LED installation.

 

 

You mean tying them together in a knot doesn't work?